Problem: $\int \left( \dfrac{3 x^{3} -4 x -2 }{x^3}\right)dx=$ $+C$
Explanation: The integrand is the quotient of two functions: $3 x^{3} -4 x -2 $ and $x^3$. Although it is tempting to take the quotient of their integrals, this would not work. $\int \dfrac{f(x)}{g(x)}\,dx\neq\dfrac{\int f(x)\,dx}{\int g(x)\,dx}$ Instead, what we should do is divide each term in the numerator by the denominator. $\begin{aligned} &\phantom{=}\int \left( \dfrac{3 x^{3} -4 x -2 }{x^3}\right)dx \\\\ &=\int\left(\dfrac{3 x^{3}}{x^3}-\dfrac{4 x}{x^3}-\dfrac{2 }{x^3}\right)dx \\\\ &=\int (3 -4 x^{-2} -2 x^{-3})\,dx \end{aligned}$ Now we can integrate using the reverse power rule, the sum rule, and the constant multiple rule for indefinite integrals. $\begin{aligned} &\phantom{=}\int \left( \dfrac{3 x^{3} -4 x -2 }{x^3}\right)dx \\\\ &=\int (3 -4 x^{-2} -2 x^{-3})\,dx \\\\ &= 3\int 1\,dx -4\int x^{-2}\,dx -2\int x^{-3}\,dx \\\\ &=3\dfrac{x^1}{1} -4\dfrac{x^{-1}}{-1} -2\dfrac{x^{-2}}{-2}+C \\\\ &=3 x + \dfrac{4}{x} + \dfrac{1}{x^2}+C \end{aligned}$ In conclusion, $\int \left( \dfrac{3 x^{3} -4 x -2 }{x^3}\right)dx=3 x + \dfrac{4}{x} + \dfrac{1}{x^2}+C$